类欧几里得算法学习笔记

如何求(sum_{i=0}^ni^{k1}lfloorfrac{ai+b}{c} floor^{k2})？((1leqslant n,a,b,cleqslant10^9,k1+k2leqslant 10))

设(f(n,a,b,c)[k1][k2])表示原式的值，则：

若(ageqslant c)，设(a=pc+q(0leqslant q<c))，则

[egin{align} &;;;;;f(n,a,b,c)[k1][k2]\ &=sum_{i=0}^ni^{k1}lfloorfrac{ai+b}{c} floor^{k2}\ &=sum_{i=0}^ni^{k1}(pi+lfloorfrac{qi+b}{c} floor)^{k2}\ &=sum_{i=0}^ni^{k1}sum_{j=0}^{k2} ext{C}_{k2}^jp^ji^jlfloorfrac{qi+b}{c} floor^{k2-j}\ &=sum_{j=0}^{k2} ext{C}_{k2}^jp^jsum_{i=0}^ni^{k1+j}lfloorfrac{qi+b}{c} floor^{k2-j}\ &=sum_{j=0}^{k2} ext{C}_{k2}^jp^jf(n,q,b,c)[k1+j][k2-j] end{align}]

只需递归计算(f(n,q,b,c))并(O(k^3))处理即可。

若(bgeqslant c)，设(b=pc+q(0leqslant q<c))，则

[egin{align} &;;;;;f(n,a,b,c)[k1][k2]\ &=sum_{i=0}^ni^{k1}lfloorfrac{ai+b}{c} floor^{k2}\ &=sum_{i=0}^ni^{k1}(p+lfloorfrac{ai+q}{c} floor)^{k2}\ &=sum_{i=0}^ni^{k1}sum_{j=0}^{k2} ext{C}_{k2}^jp^jlfloorfrac{ai+q}{c} floor^{k2-j}\ &=sum_{j=0}^{k2} ext{C}_{k2}^jp^jsum_{i=0}^ni^{k1}lfloorfrac{ai+q}{c} floor^{k2-j}\ &=sum_{j=0}^{k2} ext{C}_{k2}^jp^jf(n,a,q,c)[k1][k2-j] end{align}]

只需递归计算(f(n,a,q,c))并(O(k^3))处理即可。

若(c>a)且(c>b)，则有：

[egin{align} &;;;;;f(n,a,b,c)[k1][k2]\ &=sum_{i=0}^ni^{k1}sum_{j=0}^{lfloorfrac{ai+b}{c} floor-1}((j+1)^{k2}-[j eq0]j^{k2})\ &=sum_{j=0}^{lfloorfrac{an+b}{c} floor-1}((j+1)^{k2}-[j eq0]j^{k2})sum_{i=0}^n[j<lfloorfrac{ai+b}{c} floor]i^{k1}\ end{align}]

又

[egin{align} &;;;;;j<lfloorfrac{ai+b}{c} floor\ &Leftrightarrow j+1leqslantlfloorfrac{ai+b}{c} floor\ &Leftrightarrow jc+cleqslant ai+b\ &Leftrightarrow jc+c-b-1<ai\ &Leftrightarrow lfloorfrac{jc+c-b-1}{a} floor<i end{align}]

于是，设(T_{i,j})为(sum_{w=0}^{x}w^i)的多项式表达式的(j)次项系数

[egin{align} &;;;;;f(n,a,b,c)[k1][k2]\ &=sum_{j=0}^{lfloorfrac{an+b}{c} floor-1}((j+1)^{k2}-[j eq0]j^{k2})sum_{i=0}^n[i>lfloorfrac{jc+c-b-1}{a} floor]i^{k1}\ &=sum_{j=0}^{lfloorfrac{an+b}{c} floor-1}((j+1)^{k2}-[j eq0]j^{k2})(sum_{i=0}^ni^{k1}-sum_{i=0}^{lfloorfrac{jc+c-b-1}{a} floor}i^{k1})\ &=lfloorfrac{an+b}{c} floor^{k2}sum_{i=0}^ni^{k1}-sum_{j=0}^{lfloorfrac{an+b}{c} floor-1}sum_{p=0}^{k2-1}sum_{q=0}^{k1+1} ext{C}_{k2}^{p}T_{k1,q}j^plfloorfrac{jc+c-b-1}{a} floor^q\ &=lfloorfrac{an+b}{c} floor^{k2}sum_{i=0}^ni^{k1}-sum_{p=0}^{k2-1}sum_{q=0}^{k1+1} ext{C}_{k2}^{p}T_{k1,q}sum_{j=0}^{lfloorfrac{an+b}{c} floor-1}j^plfloorfrac{jc+c-b-1}{a} floor^q\ &=lfloorfrac{an+b}{c} floor^{k2}sum_{i=0}^ni^{k1}-sum_{p=0}^{k2-1}sum_{q=0}^{k1+1} ext{C}_{k2}^{p}T_{k1,q}f(lfloorfrac{an+b}{c} floor-1,c,c-b-1,a)[p][q] end{align}]

只需递归计算(f(lfloorfrac{an+b}{c} floor-1,c,c-b-1,a))并(O(k^4))处理即可。

当(a=0)或(n)很小时可直接计算。